How do you find the vertex, axis of symmetry, domain, range, x-intercept, and y-intercept of the equation: y=x^2+12x+35?

Question

anonymous · Answer

Vertex  is where the curve turns

anonymous · Answer

differentiate the function

hba · Answer

did u try to solve it

anonymous · Answer

no not yet, i don't understand how you find the vertex

campbell_st · Answer

the curve can be written in the form y = (x -h)^2 + k

(h, k) is the vertex, line of symmetry is x = h 

$$y = (x + 6)^2 - 1$$

the vertex is (-6, 1), line of symmetry x = -6 

domain is all real x

the range is y >= -6 

y intercept   let x = 0  then  y = 35

x intercepts    let y = 0   0 = x^2 + 12x + 35 

                                     0 = (x + 5)(x +7)

the x- intercepts are x = -5, -7

anonymous · Answer

where did you get 6 from?

anonymous · Answer

@campbell_st  how did you go from y = x^2 + 12x + 35   to    y = (x + 6)^2 − 1  ?

campbell_st · Answer

@JoshDavoll y = x^2 + 12x + 36 -1   allows you to complete the square so its
                   y = (x^2 + 12x + 36) -1

the original equation is y = x^2 + 12x + 35 which is factorised for the x intercepts...

anonymous · Answer

ahh thanks! :)

anonymous · Answer

thanks!! @campbell_st

campbell_st · Answer

the 6 comes from the completing the square

your equation is y = x^2 + 12x + 35   

by adding 1 you will get a perfect square but you also need to subtract 1

y = (x^2 + 12x + 36) - 1           is still the original equation

y = (x + 6)^2 -1                        is a factorised form of the original equation

anonymous · Answer

factor and you get
(x+7)*(x+5)

the x intercepts are  thus -7, -5   the axis of 
symmetery is midway of that    so use midpoint formula m=(x1+x2)/2 
so it is  (-7+(-5))/2= -6    
that -6 is also the x coordinant of the vertex

if you plug in -6 into  (x+7)*(x+5)=(-6+7)*(-6+5)=-1     you get  -1 for y coordinant of vertex so now you have the pair    for the coordiant of vertex  (-6, -1)  

to find domain   you ask yourself what numbers are valid that we can plug into my function.   Well it is all of them, since there is no division by zero or imaginary roots.

for range you ask yourself if i had a graph where will the y values start and where will it end. 
x^2+12x+35
since  the leading coeffecient of the highest degree term is positive it will be a "u" shaped graph. You know the vertex is (-6,-1), so it will start at -1 and go to infinity
so the   range is -1 to infinity.